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Let $\mathcal{F}$ be a family of $n$ finite sets. In this case, the family can be regarded as a multiset, since it is allowed to contain multiple instances of the same set. Let $U(\mathcal{F})$ be the universe, i.e. the union of all sets in $\mathcal{F}$.

We require that:

  1. $\mathcal{F}$ is union-closed ($\mathcal{F}$ must contain at least one instance of $A \cup B$ for any $A,B \in \mathcal{F}$);
  2. Each element in $U(\mathcal{F})$ is in at most $\lfloor (n+1)/2 \rfloor$ sets of $\mathcal{F}$ (every set is counted with its multiplicity).

Is it possible to prove or disprove that, for any possible choice of $\mathcal{F}$, there exist $A, B, C \in \mathcal{F}$ such that $A \cap B \cap C = \emptyset$? In case they must exist, is it possible to get a lower bound for the number of such unordered triplets?

Note that if we consider couples instead of triplets, it is possible to build counterexamples where there does not exist any $A, B \in \mathcal{F}$ such that $A \cap B = \emptyset$ (see here).

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  • $\begingroup$ every element belongs to at least or at most $n/2$ sets? $\endgroup$ Commented Dec 11 at 7:10
  • $\begingroup$ At most. It is the dual of the linked question. $\endgroup$ Commented Dec 11 at 7:45

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Does not a similar construction work here?

Let the elements of our universe be non-zero vectors in $K^4$, where $K$ is a finite field with, say, 239 elements. Take every hyperplane with huge multiplicity and every union of hyperplanes with multiplicity 1.

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  • $\begingroup$ Is $239$ just to pick a big number? In other words, might it work with a smaller order for $K$? $\endgroup$ Commented Dec 11 at 13:13
  • $\begingroup$ Yes, just big, to be sure that every element belongs to less than a half of hyperplanes $\endgroup$ Commented 2 days ago

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